$A$ reaction $A + B \rightarrow C$ is second order in $A$ and zeroth order in $B$. How is the rate affected when the concentration of $A$ is doubled and $B$ is halved?

The reaction $2 \ NO + Br_2 \rightarrow 2 \ NOBr$ takes place through the mechanism given below:
$NO + Br_2 \Leftrightarrow NOBr_2$ (fast)
$NOBr_2 + NO \rightarrow 2 \ NOBr$ (slow)
The overall order of the reaction is $.....$.

What is the unit of the rate constant for a $4^{th}$ order reaction?

For the gaseous reaction,$N_2O_5 \rightarrow 2NO_2 + \frac{1}{2}O_2$,the rate can be expressed as:
$-\frac{d[N_2O_5]}{dt} = K_1[N_2O_5]$
$+\frac{d[NO_2]}{dt} = K_2[N_2O_5]$
$+\frac{d[O_2]}{dt} = K_3[N_2O_5]$
The correct relation between $K_1, K_2$ and $K_3$ is:

The half-lives of two samples are $0.1 \ s$ and $0.4 \ s$. Their respective concentrations are $200$ and $50$. What is the order of the reaction?

The mechanism of reaction $2 \ NO + Cl_2 \rightarrow 2 \ NOCl$ is given as :
$(i) 2 NO \underset{k_2}{\stackrel{k_1}{\rightleftharpoons}} N_2 O _2$ (fast)
$(ii) N_2O_2 + Cl_2 \xrightarrow{K_3} 2 \ NOCl$ (slow)
The rate expression of the reaction will be :